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Consider the $3$-torus $Y=T^3$, a subset $\Sigma\subset Y$, and $\Sigma^*=Y\setminus\overline\Sigma$. We assume both $\Sigma$ and $\Sigma^*$ to be open, connected, and smoothly bounded.

I am concerned with the subsets $$ A = \ker (H^1(Y)\to H^1(\Sigma)), \quad B=\ker (H^1(Y)\to H^1(\Sigma^*))$$ of the first de Rham cohomology of $Y$. The maps are induced by the restriction maps.

My conjecture is that one of the following three statements must hold true:

  • $A=0$ or
  • $B=0$ or
  • $A=B$ and $\dim A=1$.

Can you help me prove or disprove this conjecture?

Why I expect the conjecture to be true: Honestly, I have only very crude intuitive arguments for this. If $a\in A\subset\mathbb R^3$ is nonzero, then $a$ can be written as a gradient in $\Sigma$. This means that there cannot be a closed loop $\gamma$ in $\Sigma$ "going in the direction of $a$", meaning that the fundamental group class of that loop in $Y$ (considered as an element in $\mathbb R^3$) is non-orthogonal to $a$. But the existence of such a loop is obstructed only by $\Sigma^*$, which means that there must be some sort of two-dimensional plane inside $\Sigma^*$ which is "orthogonal" to $a$. But then, given any direction $b\in\mathbb R^3$ orthogonal to $a$, we can find a closed loop in that plane (thus in $\Sigma^*$) "going in the direction of $b$". This shows that $b\notin A$.

What I have already done: I have already looked at the Mayer Vietoris sequence, but it does not seem to yield enough information. But it helps me to draw conclusions in case I already know the conjecture to be true. Indeed, denoting by $k$ the number of connected components of $\partial\Sigma$, we then know that $$\begin{align*} 1 &\ge\dim A\cap B = \dim\ker(H^1(Y)\to H^1(\Sigma)\oplus H^1(\Sigma^*)) \\ &= \dim \operatorname{im} (H^0(\partial \Sigma)\to H^1(Y)) \\ &= k - \dim\ker(H^0(\partial\Sigma)\to H^1(Y)) \\ &= k - \dim\operatorname{im}(H^0(\Sigma)\oplus H^0(\Sigma^*)\to H^0(\partial\Sigma) = k-1, \end{align*} $$ showing that $\partial\Sigma$ can have at most $2$ connected components. I have no independet proof of this result, so my second question would be if this statement is true.

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    $\begingroup$ You're on the right track with the Mayer-Vietoris sequence; have a look at the "half-lives, half-dies" principle. It tells you about the image of the restriction maps on $H^1$. Eg Lemma 3.5 of Hatcher's 3-manifold notes pi.math.cornell.edu/~hatcher/3M/3M.pdf. $\endgroup$ Dec 5, 2018 at 14:42
  • $\begingroup$ Thank you, this is interesting stuff! Can you expand a little on how this principle would help? In the Mayer Vietoris sequence there is only the difference of two such restriction maps. What makes me a bit skeptical is that my intuitive argument works in any dimension, that is, not just for the $3$-torus, whereas using the "half-lives, half-dies" principle would work only in dimension $3$ (or, any odd dimension?). $\endgroup$
    – Klaas
    Dec 6, 2018 at 17:44
  • $\begingroup$ "Half lives, half dies" is basically just an encoding of the restrictions put on maps due to Poincare duality. One way or another it's an encoding of the commutative ladder for the long exact sequence of a pair of manifolds. Generally it is most interesting around the middle dimension, which for 3-manifolds would be H^1 and H^2. $\endgroup$ May 6, 2019 at 16:49

1 Answer 1

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The following is my own attempt at an answer. I would appreciate critical proofreading.

The exact sequences of the pairs $(Y,\Sigma)$ and $(Y,\Sigma^*)$ show that $$ A=\operatorname{im}(i^*:H^1(Y,\Sigma)\to H^1(Y)), \quad B=\operatorname{im}(j^*:H^1(Y,\Sigma^*)\to H^1(Y)) $$

Take any $[u]\in H^1(Y,\Sigma)$ and $[v]\in H^1(Y,\Sigma^*)$. Here, $u$ and $v$ are closed $1$-forms on $Y$, and we can assume that $u$ vanishes of $\Sigma$ and $v$ vanishes on $\Sigma^*$. Consequently, $$ 0 = [u\wedge v]_{H^1(Y)} = [u]_{H^1(Y)} \wedge [v]_{H^1(Y)} = i^*[u]_{H^1(Y,\Sigma)} \wedge j^*[v]_{H^1(Y,\Sigma^*)} $$

We have thus shown that $A\wedge B=0$ which is but a concise formulation of the original claim.

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